Fast Auto Service provides oil and lube service for cars. It is known that the mean time taken for oil and lube service at this
garage is minutes per car and the standard deviation is minutes. The management wants to promote the business by guaranteeing a maximum waiting time for its customers. If a customer's car is not serviced within that period, the customer will receive a discount on the charges. The company wants to limit this discount to at most of the customers. What should the maximum guaranteed waiting time be
1 answer:
Answer: 19.93 minutes
Explanation:
This is the complete question
Fast Auto Service provides oil and lube service for cars. It is known that the mean time taken for oil and lube service at this garage is 15 minutes per car and the standard deviation is 2.4 minutes. The management wants to promote the business by guaranteeing a maximum waiting time for its customers. If a customer's car is not serviced within that period, the customer will receive a 50% discount on the charges. The company wants to limit this discount to at most 2% of the customers. What should the maximum guaranteed waiting time be? Assume that the times taken for oil and lube service for all cars have a normal distribution
The solution is attached below
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Answer:
Profit-maximizing price per drug treatment is $2,000
Explanation:
The "cost of production" (cost of providing all treatments) is given by the area under the cost curve
(The cost curve is the straight line C = 10Q)
It is a right triangle with one side being the quantity (Q) and the other being the cost of the last unit being produced (10Q)
So the cost of production is:
Revenue is given by P * Q = (3,000 - 10Q) * Q
Profit = Revenue - Cost of production = 3,000Q - -
To find maximum, take derivative and solve for:
3,000 - 30Q = 0 => Q = 100
Profit-maximizing quantity is 100. The price will then be P = 3,000 - 10*100 = $2,000